Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.03 m3. Assume the steam is cooled at constantvolume (i. e. the piston is held fixed in place) until the temperature reaches 200 C (callthis state 2). Then the steam is expanded isothermally until its volume is three times theinitial value (state 3). Required:
a. Determine the pressures at state 2 and 3. ans. 15.5 bar, ~10 bar
b. Determine the change in specific internal energy, u, for each of the two processes.
-389 kJ/kg, 410 kJ/kg
c. Make qualitatively correct sketches of the processes on a T-v plot.

Explanation:

Given that:

At state 1:

Pressure P₁ = 20 bar

Volume V₁ = 0.03

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; = 0.0996 / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

At state 2:

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 / kg

From temperature T₂ = 200⁰ C

Since   , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality

At temperature T₂, the specific internal energy , also

Thus,

At state 3:

Temperature

Specific volume

Thus; ,

SInce , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206

The specific internal energy at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.